This is a basic question, but I don't know the answer. Suppose $M$ is an $R$-module, considered as a complex concentrated in degree $0$. Let $F^*$ be an complex consisting of free modules. Recall that $F^*$ is called a free resolution of $M$ if there is a quasi-isomorphism $F^*\to M$.

Now suppose that $M\cong F^*$ in the derived category of $R$-modules. Does this mean that $F^*$ is a free resolution of $M$? I guess not, since maybe the map goes "in the wrong direction", but I cannot think of an example.